Cauchy sequences

1. Jun 16, 2005

steven187

hello all

I found this rather interesting
suppose that a sequence $${x_{n}}$$ satisfies

$$|x_{n+1}-x_{n}|<\frac{1}{n+1}$$ $$\forall n\epsilon N$$

how couldnt the sequence $${x_{n}}$$ not be cauchy? I tried to think of some examples to disprove it but i didnt achieve anything doing that, please help

thanxs

2. Jun 16, 2005

HallsofIvy

An obvious one: just add those differences. Let xn be $$\sum_{i=1}^n \frac{1}{n}$$. That series does not converge and so the sequence of partial sums is not Cauchy.

3. Jun 17, 2005

saltydog

I'm hesitant to challenge you Hall but feel compelled to do so at grave risk as Analysis is not my strong suite. Alright nothing is but I digress.

It looks like a Cauchy sequence to me based on the definition:

The sequence: $\{x_n}\}$ is a Cauchy sequence if given $\epsilon$ there exists N such that for all m,n$\leq$N we have:

$$|x_m-x_n|<\epsilon$$

For the sequence:

$$|x_{n+1}-x_{n}|<\frac{1}{n+1}$$

I take $\epsilon=\frac{1}{N+1}$;

Thus for all n>N:

$$|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon$$

4. Jun 17, 2005

rachmaninoff

That's true; but look closely - is that really the same thing as

?

5. Jun 17, 2005

HallsofIvy

For a given &epsilon;, you then choose a large enough N. No fair choosing N first and then finding &epsilon; so that N doesn't work!
You do remind me that just $$|x_{n+1}-x_n|< \epsilon$$ is not enough. You have to show that $$|x_n-x_m|< \epsilon$$ which is basically showing that the tail of the sum goes to 0.

6. Jun 17, 2005

saltydog

I suppose not.

$$|x_{n+1}-x_n|<\frac{1}{n+1}$$

does not imply

$$|x_m-x_n|<\frac{1}{n+1}$$

Still a little unclear. I'll work on it. Thanks guys.

Seriously, why do I even bother with differential equations anyway . . .

7. Jun 17, 2005

rachmaninoff

Simply put: for given N, your

$$| x_m - x_n |$$,

$$\mbox{ as } m \rightarrow \infty$$,

is not guaranteed to be any better than the (divergent) harmonic series:

$$| x_m - x_n | < \sum_{k=n+1}^{\infty} \frac{1}{n} \rightarrow \infty$$

8. Jun 17, 2005

saltydog

Alright, I'm interested in getting my math right.
Following Hall's example (a slight change):

Let:

$$x_n=\sum_{i=1}^{n+1} \frac{1}{i}$$

thus:

$$|x_{n+1}-x_n|=|\sum_{n=1}^{n+2} \frac{1}{i} - \sum_{n=1}^{n+1} \frac{1}{i}|=\frac{1}{n+2}$$

thus:

$$|x_{n+1}-x_n|=\frac{1}{n+2}<\frac{1}{n+1}$$

However, the partial sums of this harmonic series $\{x_n\}$ do not converge and thus cannot be a Cauchy sequence.

Now, I got a pot of spagetti to make . . .

9. Jun 17, 2005

steven187

hello all

nice stuff saltydog, i had a similar result, i think your fine with analysis, and hallsofivy your right it is an obvious one but the funny thing is i cant think of any other examples, do you know of any others?

steven

10. Jul 27, 2011

Kevin"12

hi guys i need bounded but not cauchy function can you help me asap?

11. Jul 28, 2011

alexfloo

Kevin, consider any bounded periodic function that is not constant (like, for instance, a sinusoid). It will never become and remain arbitrarily close to anything, but it does remain within fixed bounds.